1049. Counting Ones (30)

1. 原题： https://www.patest.cn/contests/pat-a-PRactise/1049

2. 思路：

3. 源码（已AC）：

#include<iostream>using namespace std;int main(void){	//freopen("in.txt", "r", stdin);	int N, sum = 0;	cin >> N;	int wgt = 1;//位权初始1，表示从个位开始		while (N / wgt != 0)	{		int left = N / (wgt * 10);//表示高位数		int cur = N / wgt % 10;//当前位值		int right = N % wgt;//低位数值		if (cur == 0)//三种情况			sum += left * wgt;		else if (cur == 1)			sum += left * wgt + right + 1;		else			sum += (left + 1) * wgt;		wgt *= 10;//位权增加	}	cout << sum << endl;	return 0;}参考自： http://m.blog.csdn.net/article/details?id=38846057